Ytukyg

I need help with this problem.

Problem. Consider the Banach space $ell^1$. Let $I$ be the identity operator and $S:ell^1 to ell^1$ be the shift operator, i.e. $S((x))_k=x_{k+1}$. Now define the family of operators $T_alpha :=I+alpha S$ for $alphainmathbb{R}.$
Determine all values of $alphainmathbb{R}$ such that the inverse of $T_alpha$ is bounded.

The notion of inverse that is used here is as an operator $T_alpha^{-1}:mathcal{R}(T_alpha)to ell^1$. Where $mathcal{R} $ denotes the range. For the inverse operator to exist only injectivity is needed.

Attempt.

The previous problem asked for values of $alpha$ such that the inverse exist. By some straightforward calculations I found that for $|alpha|leq 1$ the inverse exists. After doing this and that I found the inverse, namely:
begin{align}
T^{-1}_alpha y=left(d(1), d(2), ......right)
end{align}
where $$d(j) = sum_{kgeq j}(-alpha)^{k-j}y(k)$$

Now I split cases.

Case 1 $alpha=1$ (case $alpha=-1$ is analoguos). Take $x_n=frac{1}{n}(1,-1,...,1,-1,0,0,0)$ with $n$ nonzero elements. Clearly $Vert x_nVert =1$. We have for this $x$:
begin{align}
d_n(j)=frac{1}{n}sum_{k=j}^n(-1)^{k-j}(-1)^{k-1}=frac{n+1-j}{n}(-1)^{j+1}
end{align}
So:
begin{align}
Vert T_alpha^{-1} x_nVert = sum_{j=1}^n frac{n+1-j}{n}=frac{n+1}{2}to infty
end{align}
Hence $T_alpha^{-1}$ is not bounded for $alpha=1$. Something similar can be done for the case $alpha=-1$.

Case 2 $|alpha|<1$. I can do something similar as above, but that is too constructive and lost insight. So I thought to do it in an easier way. I thought about the following theorem that says that the existence of some $b>0$ such that for all $xinell^1$: $Vert T_alpha x Vertgeq bVert xVert$ implies the boundedness of $T^{-1}_alpha$.

I wanted to use it like this. Let $xneq 0$. One has by the reverse triangle inequality: $$frac{Vert T_alpha x Vert}{Vert xVert } geq frac{1}{Vert xVert }bigg| Vert IxVert-|alpha|Vert SxVertbigg|$$ One knows that $Vert SVert =1$, but that did not brought me anywhere.

Question. Is my solution for $alpha=1$ correct? Is there easier ways to do this exercise, expecially for the case $|alpha|<1$?

For $alpha=0$ you have $T_alpha=I$, so from now on let us assume that $alphane0$.

When $|alpha|<1$, we can calculate the inverse of $T_alpha$ explicitly via the Neumann series: namely,
$$
T_alpha^{-1}=sum_{n=0}^infty (-1)^nalpha^nS^n.
$$
So

$T_alpha$ is invertible for all $alpha$ with $|alpha|<1$.

Note that $$T_alpha=I+alpha S=alphaleft(frac1alpha,I+Sright).$$ Seeing it like this, we know that $T_alpha$ is invertible precisely when $-1/alphanotinsigma(S)$.

We have $|S|=1$; by the general result that $|lambda|>|S|$ implies that $lambdanotinsigma(S)$, we immediately get that $|1/alpha|>1$ implies $T_alpha$ invertible. This is another way to see that $T_alpha$ is invertible when $|alpha|<1$.

For any $lambda$ with $|lambda|<1$, if $x=(lambda^n) $, then $Sx=lambda x$, and so $lambdainsigma (S) $. As the spectrum is always closed, we have $$sigma (S)={lambda: |lambda|leq1}. $$ So

$T_alpha$ is not invertible if $|alpha|geq1$.